Lines and Triangles — Maths STD 9 — Question

3. 45º

Solution:

$\angle\text{AOD}=\angle\text{COB}=\theta$

$\angle\text{AOC}=\angle\text{BOD}=\phi$

Since the sum of the measures of the angles around a point is 360º,

$\angle\text{AOD}+\angle\text{COB}+\angle\text{AOC}+\angle\text{BOD}=360^\circ$

$\Rightarrow\theta+\theta+\phi+\phi=360$

$\Rightarrow2(\theta+\phi)=360$

$\Rightarrow\theta+\phi=180$

Given that $\theta=3\phi.$

So, $3\phi+\phi=180$

$\Rightarrow4\phi=180$

$\Rightarrow\phi=45$